Problem 53 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 53-56: Rewrite the expression using a radical. (a) \(4 x^{3 / 2}\) (b) \((4 x)^{3 / 2}\)

Step-by-Step Solution

Answer

(a) \(4\sqrt{x^3}\), (b) \(8x^{3/2}\).

Step 1: Identify the Format

To rewrite the expression using a radical, recognize the fractional exponent. The expression with a fractional exponent \( x^{m/n} \) is equivalent to \( \sqrt[n]{x^m} \).

Step 2: Rewrite (a) Using a Radical

The expression is \( 4x^{3/2} \). The fractional exponent \( \frac{3}{2} \) indicates that it is equivalent to \( x^{3/2} = (x^3)^{1/2} = \sqrt{x^3} \). Therefore, the expression can be rewritten as \( 4\sqrt{x^3} \).

Step 3: Simplify if Possible

Check if the expression can be simplified further. In this case, no further simplification is possible for \( 4\sqrt{x^3} \).

Step 4: Rewrite (b) Using a Radical

The expression is \((4x)^{3/2}\). Again, use the fractional exponent \( \frac{3}{2} \) to rewrite as a radical: \((4x)^{3/2} = \sqrt{(4x)^3} \).

Step 5: Simplify the Radical Expression

Simplify \( \sqrt{(4x)^3} \) by first expanding \((4x)^3 \). This gives \( 64x^3 \), so \( \sqrt{(4x)^3} = \sqrt{64x^3} \). We know \( \sqrt{64x^3} = 8x^{3/2} \), as \( \sqrt{64} = 8 \).

Key Concepts

Fractional Exponents

Understanding fractional exponents is key to working with radical expressions. A fractional exponent like \( x^{m/n} \) has two components: the numerator \( m \) and the denominator \( n \). The numerator \( m \) represents the power to which the base \( x \) is raised, while the denominator \( n \) denotes the root of the base.
Thus, \( x^{m/n} \) is equivalent to \( \sqrt[n]{x^m} \). This relationship allows you to easily switch between exponential and radical forms.
Whenever you encounter an expression such as \( x^{3/2} \), you can interpret this as the square root of \( x \) raised to the third power, hence \( \sqrt{x^3} \). This method of conversion is immensely helpful in simplifying complex expressions and solving equations.

• The numerator indicates the power.
• The denominator indicates the root.
• \( x^{m/n} = \sqrt[n]{x^m} \).

Transfer this knowledge to expressions like \( 4x^{3/2} \) by treating them as two separate parts, the constant, and the variable with a fractional exponent, rewriting them as radicals for simplicity.

Simplifying Radicals

Simplifying radicals involves breaking down the expression under the radical to its simplest form while ensuring clarity and precision.
Take an expression like \((4x)^{3/2}\), which can be rewritten as a radical: \( \sqrt{(4x)^3} \). Before proceeding, expand \((4x)^3\) to get \(64x^3\).
This allows for easier calculations and further simplification. The square root of 64 is straightforward, yielding 8, and the remaining part of the expression becomes \(x^{3/2}\). Therefore, the simplified form is \(8x^{3/2}\). Ensuring simplicity and clarity at each step is vital for avoiding errors and achieving accurate results.

• Identify potential for simplifying the radicand.
• Combine like terms under the radical for easier computation.
• Simplify constants and variables separately before combining them at the end.

Remember, the goal is achieving the simplest form while maintaining the expression's equivalence.

Exponents and Radicals

Exponents and radicals provide alternate ways to work with powers and roots in mathematics, each serving unique purposes.
An exponent expresses how many times a number multiplies itself, for example, \( x^3 \) means \( x \times x \times x \). Radicals, on the other hand, indicate the root of a number. For instance, \( \sqrt{x} \) seeks the number which, when squared, equals \( x \).
The conversion between these two forms helps simplify complex expressions and solve equations more efficiently. Equations like \( x^{m/n} \) can be switched to a radical as \( \sqrt[n]{x^m} \). This interchangeability enhances your flexibility in solving problems.

• Exponents represent repeated multiplication.
• Radicals seek the root of a value.
• The two are interconnected, offering insight into complex expressions.

Often, combining these techniques leads to simpler, more manageable mathematical problems.